Do the multiplications in $x\left(x-2\right)-\left(2x^{2}-8\right)$.
$$\frac{x^{2}-2x-2x^{2}+8}{x-2}=0$$
Combine like terms in $x^{2}-2x-2x^{2}+8$.
$$\frac{-x^{2}-2x+8}{x-2}=0$$
Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $x-2$.
$$-x^{2}-2x+8=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx+8$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-2$$ $$ab=-8=-8$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-8$.
$$1,-8$$ $$2,-4$$
Calculate the sum for each pair.
$$1-8=-7$$ $$2-4=-2$$
The solution is the pair that gives sum $-2$.
$$a=2$$ $$b=-4$$
Rewrite $-x^{2}-2x+8$ as $\left(-x^{2}+2x\right)+\left(-4x+8\right)$.
$$\left(-x^{2}+2x\right)+\left(-4x+8\right)$$
Factor out $x$ in the first and $4$ in the second group.
$$x\left(-x+2\right)+4\left(-x+2\right)$$
Factor out common term $-x+2$ by using distributive property.
$$\left(-x+2\right)\left(x+4\right)$$
To find equation solutions, solve $-x+2=0$ and $x+4=0$.
$$x=2$$ $$x=-4$$
Variable $x$ cannot be equal to $2$.
$$x=-4$$
Steps Using the Quadratic Formula
To add or subtract expressions, expand them to make their denominators the same. Multiply $x+2$ times $\frac{x-2}{x-2}$.
Do the multiplications in $x\left(x-2\right)-\left(2x^{2}-8\right)$.
$$\frac{x^{2}-2x-2x^{2}+8}{x-2}=0$$
Combine like terms in $x^{2}-2x-2x^{2}+8$.
$$\frac{-x^{2}-2x+8}{x-2}=0$$
Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $x-2$.
$$-x^{2}-2x+8=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-2$ for $b$, and $8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Divide $2$, the coefficient of the $x$ term, by $2$ to get $1$. Then add the square of $1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}+2x+1^{2}=8+1^{2}$$
Square $1$.
$$x^{2}+2x+1=8+1$$
Add $8$ to $1$.
$$x^{2}+2x+1=9$$
Factor $x^{2}+2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+1\right)^{2}=9$$
Take the square root of both sides of the equation.
